Study questions platform-wide or filter by specific tests with correct answers revealed.
Management of Acute Severe Hyperkalemia β Stepwise Approach:
| Step | Drug/Intervention | Onset | Mechanism |
|---|---|---|---|
| 1 (First priority) | IV Calcium Gluconate 10% (10 mL) or Calcium Chloride | 1β3 min | Stabilizes cardiac membrane (does NOT lower K\(^+\)) |
| 2 | IV Insulin + Dextrose (10 units regular insulin + 50 mL D50%) | 15β30 min | Shifts K\(^+\) intracellularly |
| 3 | Salbutamol nebulization (10β20 mg) | 15β30 min | Shifts K\(^+\) intracellularly via \(\beta_2\) stimulation |
| 4 | Sodium Bicarbonate (if acidotic) | 30β60 min | Shifts K\(^+\) intracellularly |
| 5 | Kayexalate / Patiromer | Hours | GI elimination of K\(^+\) |
| 6 | Hemodialysis | Immediate | Definitive removal β for refractory cases |
The principle of Unity of Command, introduced by Henri Fayol as one of his 14 Principles of Management, states that each employee should receive orders from only one superior. This prevents confusion, conflicting instructions, and breakdown of authority.
In nursing management, violating this principle leads to:
- Role confusion among staff nurses
- Conflicting priorities
- Poor accountability
- Staff stress and burnout
A Head Nurse must establish a clear chain of command to maintain effective nursing service delivery.
A force $\vec{F} = 6t^2\,\hat{i} + 4t\,\hat{j}$ acts on a particle of mass $3\text{ kg}$, initially at rest. Find its velocity at $t = 3\text{ s}$.
Acceleration: $\vec{a} = \dfrac{\vec{F}}{m} = 2t^2\,\hat{i} + \dfrac{4t}{3}\,\hat{j}$
Velocity: $\vec{v} = \displaystyle\int_0^3 \vec{a}\,dt = \left[\dfrac{2t^3}{3}\,\hat{i} + \dfrac{2t^2}{3}\,\hat{j}\right]_0^3 = \dfrac{2(27)}{3}\,\hat{i} + \dfrac{2(9)}{3}\,\hat{j} = \mathbf{18\hat{i} + 6\hat{j}}$
A test paper has 5 questions, each with three answer choices of which only one is correct. What is the probability that a student scoring purely by guessing gets 4 or more questions right?
This is a binomial distribution with $n=5$, $p=\dfrac{1}{3}$ (correct), $q=\dfrac{2}{3}$ (wrong).
$P(X \geq 4) = P(X=4) + P(X=5)$
$P(X=4) = \binom{5}{4}\left(\dfrac{1}{3}\right)^4\left(\dfrac{2}{3}\right)^1 = 5 \cdot \dfrac{1}{81} \cdot \dfrac{2}{3} = \dfrac{10}{243}$
$P(X=5) = \binom{5}{5}\left(\dfrac{1}{3}\right)^5 = \dfrac{1}{243}$
$P(X \geq 4) = \dfrac{10}{243} + \dfrac{1}{243} = \dfrac{11}{243} = \dfrac{11}{3^5}$
Sign in to join the conversation and share your thoughts.
Log In to Comment